Scalar line integrals

Scalar line integrals I

(a)

(b)

Figure 2.3: (a) Plot of a surface $z=f(x, y)$ and a curve or path in the $x y$ plane, shown in blue, given by $\mathcal{C}$ (b) The line integral represents the area of the vertical surface formed between the surface $z=f(x, y)$ and the curve in the $x y$-plane, $\mathcal{C}$.

We define the scalar line integral of a function $f$ over a curve $\mathcal{C}$ as:

such integrals are useful in engineering as they represent such quantities as total mass.

If we first consider the more familiar integral, $\int_{x_{0}}^{x_{1}} g(x) d x$, we note that the integral defines an area under some curve $g(x)$, above the $x$-axis between the points $x_{0}$ and $x_{1}$. The line integral involves a similar concept; instead of a curve $g(x)$, we have a surface, $z=f(x, y)$, as shown in Fig. 2.3. We also define a path on the $x y$-plane, given by the curve $\mathcal{C}$ in Fig. 2.3)(a). In the one-dimensional case, we want to calculate the area above the $x$-axis over some interval $\left[x_{0}, x_{1}\right]$. The line integral of a scalar field gives the surface area defined between some curve, $\mathcal{C}$ and a surface, $z=f(x, y)$. This vertical 'sheet' between $\mathcal{C}$ and $z=f(x, y)$ is shown in light blue in Fig. 2.3(b).

Suppose that we define the following parametrisation for $\mathcal{C}$ :

where the endpoints $a$ and $b$ are the endpoints of the curve $\mathcal{C}$. The parametrisation given by Eq. (2.35) can also be written as:

where $\boldsymbol{r}(t)$ is the parametrisation of curve $\mathcal{C}$. We assume that $\boldsymbol{r}(t)$ has a continuous derivative given by $\boldsymbol{r}^{\prime}(t)$ :

The magnitude of $\boldsymbol{r}^{\prime}(\boldsymbol{t})$ is determined using the Pythagorean theorem:

These ideas will become useful later on when we calculate the line integral.

To calculate the area of the vertical sheet shown in Fig. 2.3(b), we spread the 'sheet' out to obtain a flat area. The beginning corner is at $t=a$ while the ending corner is at $t=b$, as shown in Fig. 2.4. The height of this surface is the value of $z=f(x, y)$ (shown in red in Fig. 2.4). The points $x$ and $y$ correspond to those that the vector $\boldsymbol{r}$ returns when we plug in values of $t$, and $f(x, y)$ is written in terms of the vector $\boldsymbol{r}(t)$ [see Eq. (2.36)] as $f(\boldsymbol{r}(t))$. We divide the sheet area into very small rectangles of length $d s$ (see Fig. 2.4), where $s$ is the so-called arc length; $d s$ is also referred to as the line element. The height of the rectangles is given by the value of $f(\boldsymbol{r}(t))$ and their area is $f(\boldsymbol{r}(t)) d s$. The process of finding the whole area then is familiar; we integrate the quantity $f(\boldsymbol{r}(t)) d s$ over the interval between $t=a$ and $t=b$ :

where the above integral is equivalent to the definition given by Eq. (2.34).

The length $d s$ corresponds to a very small part of the curve $\mathcal{C}$ and can be represented by a straight line segment; the latter has a very small increase in the $x$-direction, $d x$, and a very small increase in the $y$-direction, $d y$. Thus $d s$ can be thought of as the hypotenuse of a triangle with sides $d x$ and $d y$, hence

Figure 2.4: On the left, we have the vertical surface from Fig. 2.3 whose area is defined by the scalar integral given by Eq. (2.34). On the right, we have this area spread out flat with the beginning and ending corners corresponding to points $t=a$ and $t=b$, respectively. The height of this area is determined by the value of the function $f(\boldsymbol{r}(t))$. Finally, very small rectangles of length $d s$ and height $f(\boldsymbol{r}(t))$ are used to determine the area of the surface.

Using Eqs. (2.38) and (2.40), we have:

Substituting Eq. (2.41) into Eq. (2.39) gives us the line integral in a form which we can integrate, as follows:

Computing a scalar line integral

Let $\boldsymbol{r}(t)$ be a parametrisation for a curve $\mathcal{C}$ for $a \leq t \leq b$. If $f(x, y)$ and $\boldsymbol{r}^{\prime}(t)$ are continuous then,

Properties:

(a) If a curve $\mathcal{C}$ is piecewise smooth, then we define:

This enables us to carry out integrals around closed curves (or loops).

(b) There may be different parametrisations for every curve but the line integral is independent of the parametrisation.

(c) The line integral is also independent of orientation. For example, suppose $\mathcal{C}_{1}$ is a curve that connects the points $(0,0)$ to $(1,1)$ while $\mathcal{C}_{2}$ is the same curve but it connects the points in the opposite direction, i.e. from $(1,1)$ to $(0,0)$. A specified direction along a path is called orientation. When $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are two parametrisations of the same curve but with opposite orientation, we write $\mathcal{C}_{2}=-\mathcal{C}_{1}$. So, we have:

(d) Finally, the ideas discussed in this section are easily generalised to threedimensional functions, i.e. $f(x, y, z)$ with $\boldsymbol{r}(t)=(x(t), y(t), z(t))$ being the parametrisation of a curve or path $\mathcal{C}$. Example 2.5 Calculate the following line integral, $\int_{\mathcal{C}} \frac{y}{x} d s, x \neq 0$; where $\mathcal{C}$ is the curve parametrised by $\boldsymbol{r}(t)=\left(t^{4}, t^{3}\right)$ for $0.5 \leq t \leq 1$.

Solution

The curve $\mathcal{C}$ is given by $\boldsymbol{r}(t)$; differentiating with respect to $t$ :

Following Eq. (2.41), the quantity $d s$ is:

We have $f(x, y)=\frac{y}{x}$ and so $f(\boldsymbol{r}(t))$ is:

which gives the integral as

Using a $u$-substitution, i.e. $u=16 t^{2}+9$ and $d u=32 t d t$, we obtain $\mathcal{I}=\frac{1}{48}\left(5^{3}-\sqrt{13}^{3}\right)$.

Scalar line integrals II

Here, we examine line integrals of functions with respect to $x$ or $y$. As before, we start with a function $f(x, y)$ and a curve parametrised by Eq. (2.35) or Eq. (2.36). Then, the line integral of $f$ with respect to $x$ is given by:

Similarly, the line integral of $f$ with respect to $y$ is given by:

We note that the integrals with respect to $x$ and $y$ given by Eqs. (2.46) and (2.47), respectively, often appear together so we have the following combined integral for two functions, say, $f(x, y)$ and $g(x, y)$ :

Example 2.6 Evaluate $\int_{\mathcal{C}} \sin (\pi y) d y+y x^{2} d x$ where $\mathcal{C}$ is a line from $(0,2)$ to $(1,4)$.

Solution

We now have a line between two points which we can write in terms of a parametrisation. To parametrise a line segment, define the two points as $\boldsymbol{r}_{0}=(0,2)$ and $\boldsymbol{r}_{1}=(1,4)$. Then the parametrisation of the line is given by:

In this example, we have:

We therefore have $x(t)=t$ and $y(t)=2+2 t$ which give $x^{\prime}(t)=1$ and $y^{\prime}(t)=2$. Using the definitions in Eqs. (2.46) and (2.47), yield the line integral as:

Note that switching the direction of the curve does not give the same result here as in the case of computing the line integral with respect to an arc length, $s$. Attempting to evaluate the line integral in Example 2.6 along the line segment in the reverse direction, i.e. from $(1,4)$ to $(0,2)$ gives $-\frac{7}{6}$. The differentials $d x$ and $d y$ may be positive or negative depending on which direction we choose to go from one point to another. Hence, if the differentials are negative, summing over negative areas to evaluate the integral, yields a negative result. We contrast this with the first case of line integrals we looked at with respect to arc length, $s$. From the definition of the arc length differential, $d s$ [see Eq. (2.40)], we can see that $d s$ is independent of the sign of the differentials $d x$ and $d y$.

We conclude this section with the following fact: if $\mathcal{C}$ is any curve, then

Following Eq. (2.49), we can have the combined form of the two integrals for two functions $f(x, y)$ and $g(x, y)$ as: